Question: Consider the equation $7\cdot e^{0.1t}=47$. Solve the equation for $t$. Express the solution as a logarithm in base- $e$. $t=$ Approximate the value of $t$. Round your answer to the nearest thousandth. $t\approx$
Explanation: The process To solve an exponential equation, we must first isolate the exponential part. Then, we can solve for the exponent by converting the equation to logarithmic form using the following equivalence: $b^t=a\,\Leftrightarrow\,\log_b{a}=t$ Isolating the exponent Let's isolate the exponent in this equation: $\begin{aligned}7\cdot e^{0.1t}&=47\\\\ e^{0.1t}&=\dfrac{47}{7}\end{aligned}$ Converting to log form and solving for $t$ If we write the above equation in logarithmic form, we get: $\begin{aligned}\log_{e}\left(\dfrac{47}{7}\right)&=0.1t\\\\ \dfrac{\ln\left(\dfrac{47}{7}\right)}{0.1}&=t\\\\ 10\cdot\ln\left(\dfrac{47}{7}\right)&=t\end{aligned}$ Note that $\log_{e}\left(\dfrac{47}{7}\right)=\ln\left(\dfrac{47}{7}\right)$. Approximating the value of $t$ Since the solution is a base- $e$ logarithm, we can plug this expression into the calculator to evaluate it. $10 \cdot \ln\left(\dfrac{47}{7}\right)\approx 19.042$ The solution is: $\begin{aligned}t&=10 \cdot \ln\left(\dfrac{47}{7}\right)\\\\ &\approx19.042\end{aligned}$